Problem 23 - Precalculus: Mathematical for Calculus

Question

The system of linear equations has a unique solution. Find the solution using Gaussian elimination or Gauss-Jordan elimination.$$\left\\{\begin{aligned} 2 x-3 y-z &=13 \\ -x+2 y-5 z &=6 \\ 5 x-y-z &=49 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solution is \(x = \frac{945}{16}\), \(y = -\frac{151}{4}\), \(z = -\frac{25}{8}\).

Step 1: Write the augmented matrix

First, write the system of equations as an augmented matrix. The equations are:\[\begin{aligned}2x - 3y - z &= 13, \-x + 2y - 5z &= 6, \5x - y - z &= 49.\end{aligned}\]The corresponding augmented matrix is:\[\begin{bmatrix}2 & -3 & -1 & | & 13 \-1 & 2 & -5 & | & 6 \5 & -1 & -1 & | & 49\end{bmatrix}.\]

Step 2: Apply Row Operations to Achieve Row-Echelon Form

We will start by making the first column below the first row zeros. Multiply the first row by 1 and add it to the second row to eliminate the -1 in the first position of the second row. Likewise, multiply the first row by -5 and add it to the third row. This gives us the new matrix:\[\begin{bmatrix}2 & -3 & -1 & | & 13 \0 & -1 & -6 & | & 19 \0 & 14 & 4 & | & -16\end{bmatrix}.\]

Step 3: Make Further Row Operations to Achieve Upper Triangular Form

Continue by eliminating the first term in the third row. Multiply the second row by 14 and add it to the third row to eliminate the 14 from the second column of the third row. The matrix becomes:\[\begin{bmatrix}2 & -3 & -1 & | & 13 \0 & -1 & -6 & | & 19 \0 & 0 & -80 & | & 250\end{bmatrix}.\]

Step 4: Back Substitution to Find Solution

Now, we have an upper triangular matrix and can perform back substitution. Begin with the third row:\[-80z = 250 \z = \frac{250}{-80} = -\frac{25}{8}.\]Back substitute into the second row:\[-y - 6(-\frac{25}{8}) = 19 \y = -19 - \frac{150}{8} = -19 - \frac{75}{4} \y = -\frac{151}{4} \\]Finally, substitute back into the first row to find x:\[2x - 3(-\frac{151}{4}) - (-\frac{25}{8}) = 13 \2x = 13 + \frac{453}{4} + \frac{25}{8} \2x = \frac{52}{4} + \frac{453}{4} + \frac{25}{8} \2x = \frac{945}{8} \x = \frac{945}{16}\]

Step 5: Write the Solution

Now we express the solution based on the calculations from back substitution. The unique solution to the system of equations is \(x = \frac{945}{16}, y = -\frac{151}{4}, z = -\frac{25}{8}\).

Key Concepts

Linear Equations

Linear equations are mathematical expressions that involve variables raised only to the first power. They do not include any exponents or square roots and often represent real-world situations such as calculating distances or determining costs. In this exercise, we have three linear equations, where each equation takes the form of a line in a three-dimensional space:

• \(2x - 3y - z = 13\)
• \(-x + 2y - 5z = 6\)
• \(5x - y - z = 49\)

Each equation involves three variables: \(x\), \(y\), and \(z\). The goal of solving these equations is to find the values of these variables that satisfy all equations simultaneously. This involves finding a unique intersection point of these three lines in space.

Augmented Matrix

An augmented matrix is a handy tool used in linear algebra to simplify the process of solving systems of linear equations. It combines the coefficients of the variables of each equation, alongside the constants from the equations, into a single matrix format. The augmented matrix for our system of equations looks like this:\[\begin{bmatrix}2 & -3 & -1 & | & 13 \-1 & 2 & -5 & | & 6 \5 & -1 & -1 & | & 49\end{bmatrix}.\]Here, the vertical line separates the matrix into two parts:

• The left side represents the coefficients of the variables \(x\), \(y\), and \(z\).
• The right side consists of the constants from each equation.

By representing the system this way, we can apply matrix operations to find the solution more efficiently.

Back Substitution

Back substitution is the process used after transforming the system of equations into an upper triangular form through Gaussian elimination. This method allows us to solve for the unknown variables starting from the last row and moving upwards. After reaching the form:\[\begin{bmatrix}2 & -3 & -1 & | & 13 \0 & -1 & -6 & | & 19 \0 & 0 & -80 & | & 250\end{bmatrix},\]we start with the third row, which only has the variable \(z\):- Solve \(-80z = 250\), giving \(z = -\frac{25}{8}\).Then, back substitute \(z\) into the second row:- Use \(-y - 6(-\frac{25}{8}) = 19\) to find \(y\).Continue by substituting \(y\) and \(z\) into the first row to find \(x\). Back substitution helps get us to the final values of \(x = \frac{945}{16}\), \(y = -\frac{151}{4}\), and \(z = -\frac{25}{8}\).

Row Operations

Row operations are the three main types of operations that can be performed on the rows of an augmented matrix to solve a system of linear equations. They include:

• Swapping two rows.
• Multiplying a row by a nonzero constant.
• Adding or subtracting the multiple of one row to another row.

These operations are crucial in transforming the matrix into row-echelon or reduced row-echelon form. Let's look at some key steps:1. To eliminate the \(-1\) in the second row of the first column, we added a multiple of the first row to the second. 2. Similarly, we added multiples of other rows to eliminate values and form zeros below the leading entries (the first non-zero number from the left in a row).Through systematic row operations, the matrix becomes simpler to handle, making it easier to apply back substitution and solve for the variables.